Amplitude, period and phase shift of can be recovered from the graph by noticing the coordinates of the peaks and troughs of the wave.
Let $y_{peak}$ and $y_{trough}$ denote the $y$-coordinates of the peaks and troughs of the wave. Then for the amplitude we have
$$ A=\frac{1}{2}\left(y_{peak}-y_{trough}\right) $$
From your graph there is no indication of the vertical scale, but let us suppose that the horizontal dashed lines are one unit apart. Then we would have $y_{peak}=2$ and $y_{trough}=-6$. This would give $A=\frac{1}{2}(2-(-6))=4$.
We can also use $y_{peak}$ and $y_{trough}$ to find the vertical shift $D$.
$$ D=\frac{1}{2}\left(y_{peak}+y_{trough}\right) $$
So, for your example, $D=\frac{1}{2}(2+(-6))=-2$.
This leaves the values of $B$ and $C$. But first, we must find the period $P$, which is straightforward.
The period P is the horizontal distance between two successive peaks of the graph. For your graph this would be a distance $P=3\pi$.
The value of $B$ is then found from
$$ B=\frac{2\pi}{P} $$
For your graph, then, we have $B=\frac{2\pi}{3\pi}=\frac{2}{3}$.
Finally we have the phase shift $\phi$.
The phase shift for the sine and cosine are computed differently. The easiest to determine from the graph is the phase shift of the cosine.
Let $x_{peak}$ denote the $x$-coordinate of the peak closest to the vertical axis. For your graph, we have $x_{peak}=0$.
$$\phi=x_{peak}\text{ for the cosine graph}$$
$$\phi=\left(x_{peak}-\frac{P}{4}\right)\text{ for the sine graph}$$
For your graph this gives phase shift $\phi=0$ for the cosine graph and phase shift $\phi=0-\frac{3\pi}{4}=-\frac{3\pi}{4}$ for the sine graph.
But for your equations, we need the value of $C$. The value of $C$ is found from the values of $\phi$ and $B$.
The equation for $C$ in terms of the phase shift $\phi$ is
$$ C=B\phi $$
So if we use the cosine function to model your graph we have $C=0$ and for the sine graph we have $C=\frac{2}{3}\cdot\left(-\frac{3\pi}{4}\right)=-\frac{\pi}{2}$.
So we have for both sine and cosine
- $A=4$
- $D=-2$
- $B=\frac{2}{3}$
For cosine, $C=0$ and for sine, $C=-\frac{\pi}{2}$.
Thus your graph can be represented by either of the two equations
$$ y=4\cos\left(\frac{2}{3}x \right)-2 $$
$$ y=4\sin\left(\frac{2}{3}x+\frac{\pi}{2}\right)-2 $$
Best Answer
We have $f(x) = \cos(\sin(x))$ and $g(x)=(1-a)\cos{2x}+a$, where $a=\frac12 (1+\cos(1))$.
Note that we can rewrite $g$ as $$g(x)=(1-a)\left(1-2\sin^2 {x}\right)+a=1-2(1-a)\sin^2{x}$$
so putting $u=\sin{x}$, we want to show that $$\cos{u} \approx 1-2(1-a)u^2$$
for $-1\le u \le 1$.
We can check this graphically, but noting that $a=0.770\ldots$ we can also see that $2(1-a)\approx \frac12$, so this is very close to the Taylor series for $\cos{u}$.
We can stop at this point, but if we want to optimise the quadratic approximation precisely over the range $-1\le u \le 1$, we can take a least squares approach. We certainly want the function to be even; so we can take $\mu+\nu u^2$ as the approximation. We want to minimise
$$\int_0^1 \left(\cos{u}-\mu-\nu u^2 \right)^2 du$$
We can explicitly calculate this integral, and differentiate with respect to $\mu$ and $\nu$ to find the optimal approximation by least squares on the interval to be $$\begin{align} \cos{u} &\approx 6 \sin(1) - \frac12 (15 \cos(1))+\left(\frac{45}{2} \cos(1) - 15 \sin(1)\right)u^2 \\ &\approx 0.997-0.465u^2 \end{align}$$
which is very close indeed to the initial approximation, which was $$g(u)=1-0.460u^2$$